Another way to better understand the FTC as the area under f(x) is to picture a line as f(x). We'll consider a simple line, y = x (can't get much simpler than that), for a better understanding of an elementary problem so the more complex ones will be better understood.
y = x
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Ok. We take the integral from a = 0 to b = 4:
4
∫ (x)dx =
0
4
(1/2)x2] =
0
(1/2)(42 - 02) =
(1/2)(16) = 8.
So 8 is the area under the graph. Now, let's look at it again. What does the area under the picture of the graph look like? Well, it looks like a triangle. So we are really just looking for the area of that triangle. Looking back to the days of basic geometry, we know the area of a triangle is (1/2)base*height. So, what is the base and the height? The base would be x-coordinate and the height would be the y-coordinate. Since we are going from x = 0 to x = 4, the length of the base is 4 - 0 or 4. Now, to find the height, we have to see where the range of f is when we go from x = 0 to x = 4. Going back to the equation of the line, when x = 0, y = 0, and when x = 4, y = 4. So, the height is also 4. Since base(or x) = 4 = height(or y), can't we say that the base*height is really just x2? So we see that the area of the triangle is (1/2)x2. When we put the value of x, 4, into the area equation, we see that the answer is, again, 8.
This is basically how the FTC works. On some AP or other standardized tests, they might give you a graph without the equation and have you find the definite integral. It will most likely be simple shapes such as triangles and rectangles (but remember, if the graph goes under the x-axis, that is negative area). With this better understanding of the FTC, these problems should be a given.
(This is, again, a very simplified exercise to show how the antiderivative works as the area under the graph. I'm sorry if I insulted any one, but I wanted to break it down severely for people that are really don't grasp the FTC.)